Russell Wilson gave the team that drafted him until midnight on Monday evening to offer up an contract extension before he cut off talks for the 2019 NFL season. Just before the clock ran out on Wilson’s self-imposed deadline, the Seattle Seahawks and their franchise quarterback reportedly came to an agreement that will make Wilson the highest paid player in the NFL.
ESPN’s Adam Schefter is reporting that Wilson and the Seahawks have come to terms on a four-year, $140 million extension, a deal that will keep Wilson in a Seahawks uniform through 2023. The deal includes a $65 million signing bonus, as well as a no-trade clause, per Schefter. Wilson’s deal tops the huge extension Aaron Rodgers got in August of 2018, both in average yearly salary and signing bonus.
Doubt had started to swirl that the two sides would be able to come to an agreement, but Wilson tweeted a video from his bed, with wife Ciara at his side, commenting on the deal being done shortly after midnight.
SEATTLE. Let’s get it. @Seahawks #GoHawks pic.twitter.com/xeWnEnUzmR
— Russell Wilson (@DangeRussWilson) April 16, 2019
“Hey Seattle. We got a deal,” Wilson says in the video.
Wilson’s agent, Mark Rodgers, issued a statement to Schefter about the extension, highlighting Wilson’s desire to continue his career in Seattle as part of the reason he “compromised to stay here,” which is certainly one way to couch your client becoming the highest paid player in the NFL.
Russell Wilson’s agent, Mark Rodgers, on the record-setting agreement that was reached at midnight PST: “At the end of the day, my guy wants to live, work, thrive in Seattle. Loves this town and it’s fans. He compromised to stay here. I respect that.”
— Adam Schefter (@AdamSchefter) April 16, 2019
With Wilson back in the fold, the Seahawks can focus on avenging their Wild Card Weekend loss to the Dallas Cowboys in the playoffs last season.